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Question Number 56878 by ajfour last updated on 25/Mar/19

f(x)= ∣∣x∣−4∣ g(x)= (√(36−x^2 ))−3 Find points of intersection.

$$\mathrm{f}\left(\mathrm{x}\right)=\:\mid\mid\mathrm{x}\mid−\mathrm{4}\mid \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\:\sqrt{\mathrm{36}−\mathrm{x}^{\mathrm{2}} }−\mathrm{3} \\ $$$$\mathrm{Find}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}. \\ $$

Answered by mr W last updated on 25/Mar/19

let t=∣x∣, 0≤t≤6 if t≤4: (√(36−t^2 ))−3=4−t (√(36−t^2 ))=7−t 36−t^2 =49−14t+t^2 2t^2 −14t+13=0 t=((7−(√(23)))/2)<4⇒ok ⇒x=±((7−(√(23)))/2) if 4≤t≤6: (√(36−t^2 ))−3=t−4 (√(36−t^2 ))=t−1 36−t^2 =t^2 −2t+1 2t^2 −2t−35=0 t=((1+(√(71)))/2)>4 ⇒ok ⇒x=±((1+(√(71)))/2)

$${let}\:{t}=\mid{x}\mid,\:\mathrm{0}\leqslant{t}\leqslant\mathrm{6} \\ $$$${if}\:{t}\leqslant\mathrm{4}: \\ $$$$\sqrt{\mathrm{36}−{t}^{\mathrm{2}} }−\mathrm{3}=\mathrm{4}−{t} \\ $$$$\sqrt{\mathrm{36}−{t}^{\mathrm{2}} }=\mathrm{7}−{t} \\ $$$$\mathrm{36}−{t}^{\mathrm{2}} =\mathrm{49}−\mathrm{14}{t}+{t}^{\mathrm{2}} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{14}{t}+\mathrm{13}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{7}−\sqrt{\mathrm{23}}}{\mathrm{2}}<\mathrm{4}\Rightarrow{ok} \\ $$$$\Rightarrow{x}=\pm\frac{\mathrm{7}−\sqrt{\mathrm{23}}}{\mathrm{2}} \\ $$$$ \\ $$$${if}\:\mathrm{4}\leqslant{t}\leqslant\mathrm{6}: \\ $$$$\sqrt{\mathrm{36}−{t}^{\mathrm{2}} }−\mathrm{3}={t}−\mathrm{4} \\ $$$$\sqrt{\mathrm{36}−{t}^{\mathrm{2}} }={t}−\mathrm{1} \\ $$$$\mathrm{36}−{t}^{\mathrm{2}} ={t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{35}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}+\sqrt{\mathrm{71}}}{\mathrm{2}}>\mathrm{4}\:\Rightarrow{ok} \\ $$$$\Rightarrow{x}=\pm\frac{\mathrm{1}+\sqrt{\mathrm{71}}}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 26/Mar/19

Thank you Sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

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